If the volume of a parallelopiped, whose coterminus edges are given by the vectors $\vec{a}=\hat{i}+\hat{j}+n \hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-n \hat{k}$ and $\vec{c}=\hat{i}+n \hat{j}+3 \hat{k}(n \geq 0)$, is 158 cu.units, then:
Correct Option: , 2
We know that the volume of parallelopiped
$=\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]$
$\left|\begin{array}{ccc}1 & 1 & n \\ 2 & 4 & -n \\ 1 & n & 3\end{array}\right|=158$
$\Rightarrow\left(12+n^{2}\right)-1(6+n)+n(2 n-4)=158$
$\Rightarrow 3 n^{2}-5 n-152=0$
$\Rightarrow 3 n^{2}-24 n+19 n-152=0$
$\Rightarrow 3 n(n-8)+19(n-8)=0$
$\Rightarrow n=8$ or $n=\frac{-19}{3}$
$\therefore \bar{a}=\hat{i}+\hat{j}+8 \hat{k}, \bar{b}=2 \hat{i}+4 \hat{j}-8 \hat{k}$ and
$\bar{c}=\hat{i}+8 \hat{j}+3 \hat{k}$
$\bar{a} \cdot \bar{c}=1+8+24=33$
$\bar{b} \cdot \bar{c}=2+32-24=10$